Which one of the following is not among the six internationally accepted classes of enzymes?

Enzymes are potent catalysts because they:

A) are consumed in the reactions they catalyze.
B) are very specific and can prevent the conversion of products back to substrates.
C) drive reactions to completion while other catalysts drive reactions to equilibrium.
D) increase the equilibrium constants for the reactions they catalyze.
E) lower the activation energy for the reactions they catalyze.

The role of an enzyme in an enzyme-catalyzed reaction is to:

A) bind a transition state intermediate, such that it cannot be converted back to substrate.
B) ensure that all of the substrate is converted to product.
C) ensure that the product is more stable than the substrate.
D) increase the rate at which substrate is converted into product.
E) make the free-energy change for the reaction more favorable.

Which one of the following statements is true of enzyme catalysts?

A) Their catalytic activity is independent of pH.
B) They are generally equally active on D and L isomers of a given substrate.
C) They can increase the equilibrium constant for a given reaction by a thousand fold or more.
D) They can increase the reaction rate for a given reaction by a thousand fold or more.
E) To be effective, they must be present at the same concentration as their substrate.

Which one of the following statements is true of enzyme catalysts?

A) They bind to substrates, but are never covalently attached to substrate or product.
B) They increase the equilibrium constant for a reaction, thus favoring product formation.
C) They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates.
D) They lower the activation energy for the conversion of substrate to product.
E) To be effective they must be present at the same concentration as their substrates.

Which of the following statements is false?

A) A reaction may not occur at a detectable rate even though it has a favorable equilibrium.
B) After a reaction, the enzyme involved becomes available to catalyze the reaction again.
C) For S → P, a catalyst shifts the reaction equilibrium to the right.
D) Lowering the temperature of a reaction will lower the reaction rate.
E) Substrate binds to an enzyme's active site.

Enzymes differ from other catalysts in that only enzymes:

A) are not consumed in the reaction.
B) display specificity toward a single reactant.
C) fail to influence the equilibrium point of the reaction.
D) form an activated complex with the reactants.
E) lower the activation energy of the reaction catalyzed.

Compare the two reaction coordinate diagrams below and select the answer that correctly describes their relationship. In each case, the single intermediate is the ES complex. (see page 190 or picture from test bank)

A) (a) describes a strict "lock and key" model, whereas (b) describes a transition-state complementarity model.
B) The activation energy for the catalyzed reaction is #5 in (a) and is #7 in (b).
C) The activation energy for the uncatalyzed reaction is given by #5 + #6 in (a) and by #7 + #4 in (b).
D) The contribution of binding energy is given by #5 in (a) and by #7 in (b).
E) The ES complex is given by #2 in (a) and #3 in (b).

Which of the following is true of the binding energy derived from enzyme-substrate interactions?

A) It cannot provide enough energy to explain the large rate accelerations brought about by enzymes.
B) It is sometimes used to hold two substrates in the optimal orientation for reaction.
C) It is the result of covalent bonds formed between enzyme and substrate.
D) Most of it is derived from covalent bonds between enzyme and substrate.
E) Most of it is used up simply binding the substrate to the enzyme.

The concept of "induced fit" refers to the fact that:

A) enzyme specificity is induced by enzyme-substrate binding.
B) enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction.
C) enzyme-substrate binding induces movement along the reaction coordinate to the transition state.
D) substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.
E) when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number ________, and the process of covalent catalysis is illustrated by the number _________.
(see picture from test bank)

A) 1; 2
B) 1; 3
C) 2; 3
D) 2; 3
E) 3; 2

The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction:

A) [ES] can be measured accurately.
B) changes in [S] are negligible, so [S] can be treated as a constant.
C) changes in Km are negligible, so Km can be treated as a constant.
D) V0 = Vmax.
E) varying [S] has no effect on V0.

Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false?

A) As [S] increases, the initial velocity of reaction V0 also increases.
B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.
C) Km is the [S] at which V0 = 1/2 Vmax.
D) The shape of the curve is a hyperbola.
E) The y-axis is a rate term with units of μm/min.

Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression:

The steady state assumption, as applied to enzyme kinetics, implies:

A) Km = Ks.
B) the enzyme is regulated.
C) the ES complex is formed and broken down at equivalent rates.
D) the Km is equivalent to the cellular substrate concentration.
E) the maximum velocity occurs when the enzyme is saturated.

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 μmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 μmol) of product to be formed?

A) 1.5 min
B) 13.5 min
C) 27 min
D) 3 min
E) 6 min

Which of these statements about enzyme-catalyzed reactions is false?

A) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration.
B) If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor.
C) The rate of a reaction decreases steadily with time as substrate is depleted.
D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction.
E) The Michaelis-Menten constant Km equals the [S] at which V = 1/2 Vmax.

For enzymes in which the slowest (rate-limiting) step is the reaction
k2
ES → P

Km becomes equivalent to:

The Lineweaver-Burk plot is used to:

A) determine the equilibrium constant for an enzymatic reaction.
B) extrapolate for the value of reaction rate at infinite enzyme concentration.
C) illustrate the effect of temperature on an enzymatic reaction.
D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.
E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by
1/V0 = Km /(Vmax[S]) + 1/Vmax.
To determine Km from a double-reciprocal plot, you would:

A) multiply the reciprocal of the x-axis intercept by −1.
B) multiply the reciprocal of the y-axis intercept by −1.
C) take the reciprocal of the x-axis intercept.
D) take the reciprocal of the y-axis intercept.
E) take the x-axis intercept where V0 = 1/2 Vmax.

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the:

A) curvature of the plot.
B) intercept on the l/[S] axis.
C) intercept on the l/V axis.
D) pK of the plot.
E) Vmax.

In competitive inhibition, an inhibitor:

A) binds at several different sites on an enzyme.
B) binds covalently to the enzyme.
C) binds only to the ES complex.
D) binds reversibly at the active site.
E) lowers the characteristic Vmax of the enzyme.

Vmax for an enzyme-catalyzed reaction:

A) generally increases when pH increases.
B) increases in the presence of a competitive inhibitor.
C) is limited only by the amount of substrate supplied.
D) is twice the rate observed when the concentration of substrate is equal to the Km.
E) is unchanged in the presence of a uncompetitive inhibitor.

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that:

A) a Glu residue on the enzyme is involved in the reaction.
B) a His residue on the enzyme is involved in the reaction.
C) the enzyme has a metallic cofactor.
D) the enzyme is found in gastric secretions.
E) the reaction relies on specific acid-base catalysis.

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition?

Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that:

F) glucose has more —OH groups per molecule than does water.
G) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.
H) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C.
I) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme.
J) water normally will not reach the active site because it is hydrophobic.

A good transition-state analog:

A) binds covalently to the enzyme.
B) binds to the enzyme more tightly than the substrate.
C) binds very weakly to the enzyme.
D) is too unstable to isolate.
E) must be almost identical to the substrate.

A transition-state analog:

A) is less stable when binding to an enzyme than the normal substrate.
B) resembles the active site of general acid-base enzymes.
C) resembles the transition-state structure of the normal enzyme-substrate complex.
D) stabilizes the transition state for the normal enzyme-substrate complex.
E) typically reacts more rapidly with an enzyme than the normal substrate.